Solving 2nd order homogenous linear ODE with a squared coefficientI am attempting to solve a differential equation of the form:d2fdx2−γ2f=0I have set up and solved the characteristic equation as:z2−γ2z+0z=0z(z−γ2)=0which is satisfied when&nbsp;z=0&nbsp;and when&nbsp;z=γ2There are two distinct roots (r1&nbsp;and&nbsp;r2) hence the general solution should be of the formf(x)=Aer1x+Ber2xIn this case:f(x)=Ae0×x+Beγ2x&nbsp;=A+Beγ2xApparently, this solution is incorrect as in the answers it is given as:f(x)=Aeγx+Be−γxI would like to know where I went wrong to give me the incorrect solution

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Solving 2nd order homogenous linear ODE with a squared coefficientI am attempting to solve a differential equation of the form:d2fdx2−γ2f=0I have set up and solved the characteristic equation as:z2−γ2z+0z=0z(z−γ2)=0which is satisfied when&amp;nbsp;z=0&amp;nbsp;and when&amp;nbsp;z=γ2There are two distinct roots (r1&amp;nbsp;and&amp;nbsp;r2) hence the general solution should be of the formf(x)=Aer1x+Ber2xIn this case:f(x)=Ae0×x+Beγ2x&amp;nbsp;=A+Beγ2xApparently, this solution is incorrect as in the answers it is given as:f(x)=Aeγx+Be−γxI would like to know where I went wrong to give me the incorrect solution

bobthemightyafm · Accepted Answer

Without appealing directly to the characteristic equation, you can also proceed as followsf″−γ2f=0&amp;nbsp;⟺(f″−γf′)+(γf′−γ2f)=0&amp;nbsp;&amp;nbsp;&amp;nbsp;⟺(f′−γf)′+γ(f′−γf)=0&amp;nbsp;&amp;nbsp;&amp;nbsp;⟺g′+γg=0where&amp;nbsp;g=f′−γf.

kweefomucy9 · Accepted Answer

Solve is as follows:d2fdx2−γ2f=0↔(d&amp;nbsp;dx&amp;nbsp;+γ)(d&amp;nbsp;dx&amp;nbsp;−γ)fTherefore,&amp;nbsp;{(ddx+γ)u=0u=(ddx−γ)fAnswer:{u=c1eγx(ddx−γ)f=c1e−γx→f=c2eγx−c1e−γx2γ=c2eγx+c3e−γx

Solving 2nd order homogenous linear ODE with a squared coefficient I am attempting to solve a di

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